# Paradox with elementary submodels of the constructible tower

This is an argument I thought up after a class on combinatrical properties of the model $\textbf{L}$. Our course is about set theory, not logic, so this paradox desn't seem relevant in its context. Can you help me figure out where I got it wrong?

The constructible heirarchy of sets is a series $L_{\alpha}$ that is defined for all ordinal numbers $\alpha$. The important properties for my argument are:
1. $L_{\alpha}$ is transitive for every $\alpha$
2. If $\alpha < \beta$, then $L_{\alpha}\subset L_{\beta}$
3. The transitive collapse (aka Montowski collapse) of every elementary submodel $M \prec L_{\alpha}$ is $L_{\beta}$ for some $\beta$
4. $L_{\omega_{1}}$ satisfies "every set is countable" and $L_{\omega_{2}}$ does not
5. $L_{\alpha}$ is coutable iff $\alpha$ is countable

So, we take an countable elementary submodel (CESM) $M_{1} \prec L_{\omega_{1}}$, and look at its transitive collapse, $L_{\alpha_{1}}$ for some countable $\alpha_{1}$. We then take an CESM $M_{2} \prec L_{\omega_{2}}$ that contains $L_{\alpha_{2}}$, and collapse it to get $L_{\alpha_{2}}$ with countable $\alpha_{2}$. Then the same procedure yields a model $L_{\alpha_{3}} \supset L_{\alpha_{2}}$ that has an elementary embedding into $L_{\omega_{1}}$. We generate an infinite series, switching between modelling $L_{\omega_{1}}$ and $L_{\omega_{2}}$.

The limit $$L_{\alpha}=L_{\lim_{n<\omega}\alpha_{n}}=\bigcup_{n<\omega}L_{\alpha_{n}}$$ is then the union of both subseries $\{L_{\alpha_{n}}\}_{n=1,3,\ldots}$ and $\{L_{\alpha_{n}}\}_{n=2,4,\ldots}$. But a union of a series of elementary submodels is itself an elementary submodel, since it is a direct limit. In particular $L_{\alpha}$ should be elementary equivalent to both $L_{\omega_{1}}$ and $L_{\omega_{2}}$. This is impossible because of property (4), namely there is a statement true in one and not in another.

Where did I go wrong in my reasoning? All kinds of tips are appreciated...

AKG
Homework Helper
"The union of elementary submodels is itself an elementary submodels" is only true when those submodels are elementary substructures of one another. This means more than just the fact that $L_{\alpha_n} \subset L_{\alpha_{n+2}}$ and $L_{\alpha_n} \equiv L_{\alpha_{n+2}}$, it requires that the inclusion map is the elementary embedding. This would be the case if the $L_{\alpha_n}$ were elementary substructures of their respective $L_{\omega_i}$, but they're generally not. You do get that they're elementarily equivalent subsets of their respective $L_{\omega_i}$, but the inclusion maps are not typically the elementary embeddings. The embeddings here are the inverses of the Mostowski collapses.

Nothing I said above proves that the inclusion maps aren't elementary embeddings, although nothing you said proves that they are. But here's why they're definitely not. Let's take for example $L_{\omega_2}$. For the same that GCH holds in L, we know that $L_{\omega_2}$ thinks there is exactly one cardinal after $\omega$, and that this cardinal is $\omega_1$. If $L_{\alpha} \equiv L_{\omega_2}$ with $\alpha$ countable, then this structure will also think there's a unique cardinal after $\omega$, but it won't think $\omega_1$ is it.

I see what you mean...

So, it seems that for every ordinal $\alpha$, the set $\{\delta < \omega_{1} \mid L_{\delta} \prec L_{\alpha}\}$ is closed w.r.t taking limits. I thought about it some more and it's not hard to see this set is unbounded for $\alpha = \omega_{1}$, since for each $\beta < \omega_{1}$ you can find find some $L_{\beta'}$ with $\beta<\beta'<\omega_{1}$ that is closed w.r.t Skolem functions of $L_{\omega_{1}}$. This natually means that the set of δs can't be unbounded for $|\alpha|>\aleph_{1}$.

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AKG
Homework Helper
I see what you mean...

So, it seems that for every ordinal $\alpha$, the set $\{\delta < \omega_{1} \mid L_{\delta} \prec L_{\alpha}\}$ is closed w.r.t taking limits. I thought about it some more and it's not hard to see this set is unbounded for $\alpha = \omega_{1}$, since for each $\beta < \omega_{1}$ you can find find some $L_{\beta'}$ with $\beta<\beta'<\omega_{1}$ that is closed w.r.t Skolem functions of $L_{\omega_{1}}$.
Correct. Another simple argument is to do the following construction:

• $X_0 = Hull(\{\beta\},L_{\alpha})$
• $\beta_0 = \min\{\gamma : X_0 \subset L_\gamma\}$
• $X_{n+1} = Hull(L_{\beta_n},L_\alpha)$
• $\beta_{n+1} = \min\{\gamma : X_n \subset L_\gamma\}$
• $L_{\beta'} = \bigcup X_n$
Here $Hull(X,M)$ denotes some Skolem Hull of the set X in the structure M. Since we're working with L, there's a canonical choice for such hull.
This natually means that the set of δs can't be unbounded for $|\alpha|>\aleph_{1}$.
Indeed, the set of such $\delta$'s is empty!